On 11/02/2013 08:55 PM, Virgil wrote:> In article <l53tef$29v$1@dont-email.me>,> "Michael F. Stemper" <michael.stemper@gmail.com> wrote:>>> On 11/02/2013 04:38 PM, Virgil wrote:>>> In article <685bd338-7570-426a-b2cb-07089fa9693f@googlegroups.com>,>>> PotatoSauce <kiwisquash@gmail.com> wrote:>>>>>>> In practice, 0^0 = 1 works just fine.>>>>>> Unless one wants 0^X to be continuous at X = 0.>>>> Which is normally not a requirement when working in the>> naturals -- the context of this discussion. I don't think>> that anybody's said that 0^0 can or should be defined for>> the reals.>> But why should one have to have different definitions for integers and> reals?

Because when you define the naturals as the finite cardinals, thedefinition of exponentiation gives 0^0 = 1 without any specialhandling.

Because the binomial theorem works without special cases if youdefine 0^0 = 1 in the naturals.

Because it's an empty product.

Because the power rule for differentiation doesn't need anyspecial cases if you define 0^0 = 1 in the naturals.

Because Knuth said so.

-- Michael F. StemperOutside of a dog, a book is man's best friend.Inside of a dog, it's too dark to read.